Chemical Composition

CHAPTER 8

Chemical Composition CHAPTER ANSWERS 1.

2.

100 washers X

0.110 g 1 washer

=

11.0 g (assuming 100 washers is exact).

100. g x

1 washer 0.llOg

500. g x

1 cork 1.63 g

500.gx

1 stopper =ll6stoppers 4.31g

=

=

909 washers

306.7 = 307 corks

1 kgofcorkscontains Il000g x

cork= 613.49=6l3corks 1.63g)

613 stoppers would weigh [613 stoppers

4.31 g 1 stopper

2644 g = 2640 g

The ratio of the mass of a stopper to the mass of a cork is (4.31 g/ 1.63 g). So the mass of stoppers that contains the same number of stoppers as there are corks in 1000 g of corks is l000gx

4.31 1.63g

3.

The atomic mass unit (amu) is defined by scientists to more simply describe relative masses on an atomic or molecular scale. One amu is equivalent to 1.66 x 1 0_24 g

4.

The average atomic mass takes into account the various isotopes of an element and the relative abundances in which those isotopes are found.

5. 6.941 amu 1 Li atom

a.

278 Li atoms x

b.

lxlO 6 Catomsx

c.

5 x 10 2 Na atoms x

d.

I Cd atom x

-

=

1930 amu

12.01 amu =1.2OlxlO 7 amu=lxlO 7 amu I C atom 22.99 ainu I Na atom

112.4 amu I Cd atom

=

1.150 x 10 27 amu = I x 10 27 amu

112.4 amu

Copyright © Houghton Mifflin Company. All rights reserved,

________

124

Chapter8 Chemical Composition 200.6 1 Hg atom

e.

6.022 x 10 Hg atoms x

a.

40.08 amu Ca x

b.

919.SalnuWx lWatorn 183.9 aniu

c.

549A aniu Mn

d.

6345amu1x

e.

2O72amux 1PbatomlOPbms 207.2 amu

1.208 x 10 amu

6.

7.

x

1 Caatom = 1 Ca atom. 40.08 amu

lMn atom = 10 Mn atoms 54.94 amu

11 atom =50latoms 126.9 aniu

One sodium atom has a mass of 22.99 amu. 124 sodium atoms would weigh: 124 atoms x 22.99 amu latom

344.85 amu represents: 344.85 x 8.

°Z

=

2851 amu

15, atoms.

One tin atom has a mass of 118.7 amu.

4

A sample contatnmg 35 tm atoms would weigh: 35 x 118.7 amu = 4155 amu I atom -

9.

itin atom 2967.5 amu of tin would represent: 2967.5 amu x 118.7 amu Avogadro’s number (6.022 x 1023 atoms; 1.00 mol)

10.

3 x Avogadro’s number (3 x 6.022 x 1023

11.

molarmasses:Na,22.99g;K,39.lOg 11.50 gNax

25 tin atoms.

1.807 x lOu, 3.00 mol)

1 22.99g

0.5002 mol Na x 0.5002molKx 12.

=

=

6.033 X 10 1 mol Na

3.0 12 x 1023 atoms

I molK

32.00 g of 02 (the molar mass of 02) contains the same number of atoms as 28.02 g of N 2 (the molar mass of N ). Each quantity represents Avogadro’s number of its respective molecules. 2

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1

Chapter 8: ChemIcal ComposItion

13.

125

The ratio of the atomic mass of H to the atomic mass Qf N is (1.008 amu/14.01 amu), and the mass of hydrogen is given by 7.00 g N x

1.008 amu = 0.504 g H. 14.01 amu

14. The ratio of the atomic mass of Co to the atomic mass ofF is (58.93 amu/19.00 amu), and the mass of cobalt is given by 57.Ogx

58.93 amu 19.00 ainu

177gCo.

-

15.

The mass of a magnesium atom equals 3.82

16.

14.01 g of nitrogen atoms contains 6.022 is given by lNatomx

17.

14.01 g 6.022 < 1023 atoms

x

x

10 —23 g

4. amu g 22.99 amuNa

=

4.04

N atoms; therefore the mass of one

2.32x10.23 g

1 molofHeatoms=4.003g 4molofHatomsx

imol

4 mol of H atoms has a mass slightly larger than I mol of He atoms. 18.

0.50molOatomsx 4molHatomsx

1 mol

1.OO8gH 4gH 1 mol

Half a mole of 0. atoms weighs more than 4 moles of H atoms. 19. 1 mol 74.92 g

a.

21.5OgAsx

b.

9.lO5gPx

c.

0.O5l52gBax

d.’ 43.l5gCx

0.287OmolAs

1 mol =0.2940mo1P 30.97 g 1 mol =3.752x 10mo1Ba 137.3 g

1 mol 12.01 g

e.

26.02 g Cr x

I mci 52.OOg

f.

1.95lgPtx

1 mci 195.1 g

S

3.593molC

=

0.5004 mcI Cr

0.Ol000molPt

CopyrIght © Houghton Mlfflln Company. All rIghts reserved.

10 —23 g

‘‘I

!cirt

nitrogen

atom

____ ____ _____ _____ _____ _= _ __=

126

Chapter 8: Chemical Composition

1 mol 19.OOg

g.

0.000375 g F x

a.

imol 66.50 g F x 19.OOg

b.

401.2mg Hg x

=

1.97 x i0 mol F

20. =

3.500 mol ofF atoms

1 mmol 200.6mg

1 mol 28.09g

2.000 mmol Hg (1 mmol

c.

84.27 g Si x

d.

1 mol 48.78 g Pt x 195.1 g

e.

2431 g Mg x

f.

47.97 g x

a.

6.94lgLi 0.251 molLi x ImolLi

b.

g 2 9 . A 4 0 l= 6 8 7 : 1.SlmolAlx 1 1 mol Al

imol 24.31 g

mol 95.94g

=

1/1000 mol)

3.000 mol Si

=

=

=

0.2500 mol Pt

=

100.0 mol Mg

0.5000 mol Mo

21.

o 2 molPb x

=

1.74 gLi

207.2 gPb 1 mol Pb

c.

8.75 x

d.

l2SmoICrx 52.00 g Cr 3 =65O g C xlO r lmolCr

e.

55.85 gFe 3 4.25 m olF xlO ex =237x10 gFe imol

f.

0.000105 mol Mg x 24.31 gMg lmolMg

S

18.1 gPb

2 55 x i0 g Mg —

22. a.

1.76x 10 3 molCsx. 132.9g =234x 10’ gCs lmol

1,.

0.0125 mol Ne x

c.

207.2 g 3 5.2 m olP 9xl bx O =1 lOx lO gPb 6 imol

20.18 g 1 mol

0.252 g Ne

Copyright © Floughton Mifflin Company. All rights reserved.

Chapter 8: Chemical Composition 22.99 = 80 lO • 2 gNa 5 I mol

d.

0.00000l22molNax

e.

5.51 millimol As x

f.

8.72mo1Cx 105gC 120 Imol

a.

1.50 g Ag

b.

0.0015 mol Cu x

c.

0.0015 g Cu

d.

g 3 2.OOkg=2.OOxlO

74.92 g 1 mol

=

413 mg = 0.4 13 g As

23.

2.00 x e.

6.022 x 1023 Ag atoms 107.9 gAg

8.37 x 1021 Ag atoms

6.022 x 1023 Cu atoms Imol

6.022 x 1023 Cu atoms 63.55 g Cu

g Mg x

6.022

=

9 Ox 10 20 Cu atoms

1.4 x i0’ Cu atoms

1023 Mg atoms 24.31 gMg

=

4.95 X 1025 Mg atoms

1.0000z28.35g 2.34ozx

28.35 g 6.022 x 1023 Ca atoms x =997x1O 23 Ca atoms 1.000oz 40.08 gCa 6.022 x 1023 Caatoms x l0 352 Caatoms 22 40.08 g Ca

f.

2.34gCax

g.

2.34 mol Ca

a.

22.99 amu 425 Na atoms x 1 Na atom

6.022 x 1023 Ca atoms = 1.41 X 1024 Ca atoms 1 molCa

24.

Na

atoms

x

9.77 x i0 3 amu

22.99 gNa = 1.62 x 1020 g 6.022 x 1023 Na atoms

b.

425

c.

425 mol Na

22.99 g Na imolNa

d.

425 mol Na

6.022 x 1023 Na atoms 1 mol Na

e.

425 gNa

=

x

6.022 x 1023 Na atoms 22.99 gNa

Copyright © Houghton Mifflin Company. All rights reserved.

g

=

=

2.56 x 10 sodium atoms

111 x 1025 sodium atoms

127

128

Chapter 8: Chemical Composition VV

f.

1 molNa 425 gNa x 22.99 gNa

g.

425 g Na

24.31 g Mg 22.99 gNa

=

18.5 molNa

449 g Mg

25.

molar mass

26.

The molar mass is calculated by summing the individual atomic masses of the atoms in the formula. V

27. a.

, chromium(ffl) oxide 3 0 2 Cr

V

mass of 2 mol Cr=2(52.00 g)= 104.00 g mass of 3 mol 0 = 3(16.00 g) = 48.00 g molar mass of Cr 3 0 2 b.

=

(104.00 g

+

V

48.00 g) = 152.00 g

, copper(II) nitrate 2 ) 3 Cu(N0 mass of 1 mol Cu = 63.55 g = 63.55 g mass of 4 mol N = 4(14.01 g) = 56.04 g mass of 6 mol 0

6(16.00 g)

=

mass of 6 mol 0 d.

V

96.00 g

molar mass of Cu(N0 2 = (63.55 g ) 3 , tetraphosphorus hex(a)oxide 6 0 4 P mass of 4 mol P

V

+

56.04 g + 96.00 g) = 215.59 g V

4(30.97 g) = 123.9 g =

6(16.00) g = 96.00 g

molar mass of P 6 = (123.9 g + 96.00 g) 0 4 , bismuth(III) oxide 3 0 2 Bi

=

219.9 g V

mass of 2 mol Bi = 2(209.0 g) = 418.0 g mass of 3 mol 0 = 3(16.00 g) = 48.00 g molar mass of Bi 3 0 2 e.

f.

=

(418.0 g

+

48.00 g) = 466.0 g

, carbon disulfide 2 CS mass of 1 mol C

=

mass of 2 mol S

=

12.01 g = 12.01 g 2(32.07 g)

molar mass of CS 2 = (12.01 g , sulfurous acid 3 HSO

=

64.14 g

+

64.14 g)

=

76.15 g

V

mass of2 molH=2(1.008 g)= 2.016 g massofl molS=32.07g=32.07g mass of 3 mol 0

=

3(16.00 g)

=

48.00 g

molar mass of 3 S0 = (2.016 g + 32.07 g 2 H

+

48.00 g)

82.11 g

Copyright © Houghton Mifflin Company All rights reserved,

Chapter 8: ChemIcal Composition

28. a.

carbon monoxide mass of I mol C = 12.01 g massoflmolO=16A30g molarmassofCO=(12.01 g+ 16.OOg)=28.01 g

b.

sodium carbonate mass of 2 mol Na = 2(22.99 g) = 45.98 g massoflmolC=12.Olg mass of 3 mol 0 = 3(16.00 g) =48.00 g molar mass of 3 CO 2 Na

c.

=

(45.98 g + 12.01 g + 4800 g) = 105.99 g

iron(II[) nitrate/ferric nitrate massoflmolFe=55.85g mass of 3 mol N = 3(14.0 1 g = 42.03 g

mass of 9 mol 0

9(16.00 g) = 96.00 g

molá mass of Fe(N0 ) = (55.85 g + 42.03 g + 96.00 g) = 241.88 .g 3 d.

hydrogen iodide

mass of ImolH= 1.008 g mass of lmoII= 126.9 g mölarmassofH[= 127.9g e.

sulfur trioxide mass of I mol S = 32.07 g mass of 3 mol 0=3(16.00 g) = 48.00 g molar mass of SO 3 = (32.07 g + 48.00 g) = 80.07 g

29. a.

2 BaC1 mass of I molBa= 137.3 g mass of 2 mol Cl = 2(35.45 g) = 70.90 g

b.

màlar mass of BaC1 2 = (137.3 g + 70.90 g) = 208.2 g ) 3 A1(N0 mass of 1 mol Al = 26.98 g mass of 3 mol N = 3(14.01 g) = 42.03 g mass of 9 moL 0 = 9(16.00 g) = 96.00 g molar mass of A1(N0 ) = (26.98 g + 42.03 g 3

Copyright © Houghton MIfflin Company. API rights reserved.

+

96.00 g) = 213.0 g

129


130 c.

2 FeC1 mass of 1 mol Fe = 55.85 g mass of2 mol C1= 2(35.45 g)= 70.90 g molar mass of FeC1 2 = (55.85 g + 70.90 g) = 126.75 g

d.

2 S massoflmolS=32.07g mass of2 molO

=

-

2(16.00 g)=32.00 g

molar mass of SO 2 = (32.07 g e.

32.00 g)

+

=

64.07 g

) 0 3 H 2 Ca(C mass ofi molCa=40.08g mass of 4 mol C

=

4(12.0 1 g) = 48.04 g

mass of 6 mol H = 6(1.008 g) = 6.048 g mass of 4 mol 0

=

4(16.00 g) = 64.00 g

molar mass of Ca(C ) = (40.08 g + 48.04 g + 6.048 g 0 3 H 2

+

64.00 g)

=

158.17 g

30. a.

aluminum fluoride mass of 1 mol Al = 26.98 g mass of3 moiF

I

=

3(19.00 g)= 57.00 g

molar mass of A1F 3 b.

=

(26.98 g

+

57.00 g) = 83.98 g

sodium phosphate mass of3 molNa=3(22.99 g) = 68.97 g mass ofl molP=30.97g mass of4molO=4(16.00 g)=64.OOg molar mass = 4 P 3 ofNa O (68.97 g + 30.97 g

c.

+

64.00 g) = 163.94 g

magnesium carbonate mass of 1 mol Mg = 24.31 g mass of 1 mol C = 12.01 g mass of 3 mol 0

=

3(16.00 g)

=

48.00 g

molar mass of MgCO 3 =(24.31 g + 12.01 g + 48.00 g)

=

84.32 g

Copyright © Houghton Mifflin Company All rights reserved


d.

131

lithium hydrogen carbonate/lithium bicarbonate mass of imol Li = 6.94 1 g massofi molH=1.008g

•

massoflmolC=12.Olg .

mass of 3 mol 0 e.

=

3(16.00 g) = 48.00 g

molar mass of LiHCO 3 = (6.941 g ÷ 1.008 g + 12.01 g + 48.00 g) = 67.96 g chromium(ffl) oxide/chromic oxide mass of 2 mol Cr = 2(52.00 g) = 104.0 g mass of 3 mol 0 = 3(16.00 g)

48.00 g

molar mass of Cr 4 = 152.0 g 0 2 31.

a.

2 molarmassof = 46.O1 NO g; x 2 0.0214gN0

b.

.mol =4.65x10molNO 46.Olg

molarmassofCu(NQ= 187.6g 1 mol 187.6g

1.56.g Cu(N0 )a x 3 c.

1 mol 76.15g

0.0324 mol

1 mol 342.2 g

molar mass of PbC1 2 2.99 g x

f.

=

molar mass 4 (S0 = 342.2 g 2 ofAl 3 )

5.04 g 4 (S0 x 2 A1 3 ) e.

3 mol CuNO 8.32.x1W 2 ) 3

=

2 molarmassof = 76.15g CS 2.47 g CS 2x

• d.

21.4mg=0.0214g

1 mci 278.lg

=

=

=

0.0 147 mci 4 (S0 2 A1 3 )

278.1 g

0.0 108 mol PbCI 2

molar mass of CaCO 3 = 100.09 g 62.4 g CaCO 3

X

1 mol = 0.623 mol CaCO 3 100.09g

32. a.

molar mass NaC1 = 58.44 g; 52.1 mg = 0.0521 g 1 mol 0.0521 g x 58.44g

=

8.92 x 10 mol

Copyright 0 Houghton Mifflin Company. All rights reserved.

132


b.

molar mass MgCO 3 io.sgx)mol 84.32g

c.

1 mol = 0.0392 mol l0l.96g’

molar mass of Fe 3 0 2 mol 159.7 g

24.1 g x e.

159.7 g

0.151 mol

•1

milhirnolar mass of 3 CO = 73.89mg 2 Li 125 mgx

f.

0.125mo1

molar mass A1 3 = 101.96 g 0 2 4.00 g x

d.

84.32 g

immol =1.ó9mmol=1.69x 1(T moI 3 73.89 mg

molar mass of Fe = 55.85 g; 2.25 kg = 2250 g L4I

2250gx imol =40.3mol 55.85g

A

,_S

33. a.

molar mass MgC1 2 mol 95.21g

41.5 g x 5.

mol = 4.52 x 29.88g

f.

mol

4.52 mmole

molar mass Cr= 52.00 g; 1.21 kg= 1210 g =

23.3 mol

molar mass HSO 4 = 98.09 g 62.5 g x

e.

0.436 mol

=

1 mol 1210 g x 52.OOg d.

95.21 g

molar mass Li 0 = 29.88 g; 135 mg = 0.135 g 2 0.135 g x

c.

=

mol 98.09g

=

0.637 mol

mola = 6 H 78 rmas .llg sC 1 mol 42.7 g x 78.llg

=

0.547 mol

molar mass H 0 2

=

34.02 g

1 mol 135 g x 34.02g

=

3.97 mol

Copyright © Houghton Muffin Company. All rights reserved.


34. &

molar mass of 4 PO 2 NaH 4.26 x 1 0 g x

b.

c.

=

120.0 g 3.55 x 1W 5 mol

99.00 g

1 mol =5.26mo1 99.00g

molarmassofFe=55.85g 151 kg x

ci.

1 mol 120.Og

molar mass of CuCI 521gx

=

l000g imol x =2.70x10 3 mol 1kg 55.85g

molar mass of SrF 2 8.76 g x

mol 125.6g

=

=

125.6 g

0.0697 mol •1

e.

molar mass of Al .1.26 x i0 g x

26.98 g

1 mol = 47 mol 26.98g

35. a.

molar mass AIC1 3

=

133.3 g

g 3 l 1 = 3 67 3 1.25 molx . lmöl b.

molar mass NaHCO 3 3.35 moix

o.

ci.

84.01 g

84.01 g 281 g imol

molar mass HBr 4.25 mmol

=

=

80.91 g

80.91 mg 1 mmol

=

344 mg

molar mass CO 2

=

=

0 312 g

44.01 g

44.01 g 0.00 104 mol x I mol f.

0.344 g

molartnassU=238.Og 238.Og 1.31 x i0 mol x imol

e.

=

0.0458 g

molarmassFe=55.85 g 55.85 1.49x 2 lO m olFex 1 mol

=8.32x 10 g 3


133

134

Chapter 8: Chemical Composition •V •

36.

;.

a.

•

molarmassofCO=28.01 g 28.01 g = 0.132 g 1 mol

0.00471 mol b.

molar mass of AuC1 3

303.4 g

=

303.4 g 1 mol

1.75 x 10 mol AuCI 3x c.

molar mass of FeCI 3

228 mol FeC1 3 d.

<

3.70 x l0 g

molar mass of LiCl

212.3 g I mol

=

2.98 millimol = 0.00298 mol

=

0.633 g

42.39 g 42.39 g I mol

2.71 x 10 mol LiCI x f.

=

molar mass of 4 P0 = 212.3 g; 3 K <

5.31 x 10 g

162.2 g

162.2 g 1 mol

0.00298 mol 4 P0 3 K e.

=

=

=

0.115 g

3 molarmassof = NH 17.03 g 6.55 molNH x 3

17.03g =112g I mol

V V.

V •

V

V

•V

V

37.

V

a.

molar mass of C 0 = 46.07 g 6 H 2 0.251 moix

b.

V,

molar mass of AuC1 3

VVV,

44.01 g 55.5 g CO 2

303.4 g

303.4 g 1 mol

=

0.282 g AuC1 3

molar mass of NaNO 3 = 85.00 g 7.74 moi x

e.

=

44.01 g I mol

9.31 x I 0 mol d.

VV

1 mol

molar mass of CO 2 1.26 mol x

c.

V

85.00 g = 658 g NaNO 3 1 mol

molar mass of Fe = 55.85 g 0.000357 mol

<

55.85 g 1 mol

=

0.0199 g Fe

Copyright © Houghton Miffhn Company. All rights reserved.


135

38. a.

molar mass NaOC1

74.44g =0.313g Imol

0.00421 molx b.

= 139.3 g 2 molar mass BaH 139.3g imol

0.998 mol x c.

74.44 g

=

139 g

=83.98g 3 molarmassAlF 83.98 g 1.99 x 10_2 mol x 1 inol

d.

molar mass MgC1 2 0.Ll9molx

e.

f.

1,67 g

95.21 g

95.21 g =11.3g I mol

molar mass Pb 225 mol x

=

=

207.2 g

=

207.2 g =4.66 x 1 mol

molar mass CO 2

=

g

44.01 g

44.01 g 0.10! mol x I mol

4.45 g

39. a.

4.75 millimol

=

0.00475 mol x b.

6.022 x 1023 molecules Imol

molar mass of PH 3 4.75 g

c.

0.00475 mol

=

33.99 g

6.022 x 1023 molecules 33.99 g

molar mass of Pb(C ) 0 3 H 2 1.25x10 g 2 x

2 86 x 10 21 molecules

=

=

8.42 x 1022 molecules

325.3 g

6.022 x 1023 formula units 325.3g

231x10 19 formula units

6.022 x l0 formula units =753x 1021 formulaunits •lmol

d.

1.25x lO rnolx 2

e.

If the sample contains a total of 5.40 mol of carbon, then because each benzene contains six carbons, there must be (5.40/6) 0.900 mol of benzene present. =

0.900molx

6.022 x 1023 molecules =54’xlO 23 molecules Imol ‘


136


40. 6.022 x 1023 molecules 74 = 3.84 x 10 molecules CO 1 mol

a.

6.37 mol CO x

b.

molar mass of CO=28.01 g 6.37gx

c.

d.

6.022 x 1023 molecules =1.37x10 23 moleculesCO 28.01 g

molar mass of H 0 2

=

18.02 g

2 62 X 10 g

6 022 x 1023 molecules 18 02 g

2.62x10gx

6.022 x 1023 xlO 58 molecu lesl ls molec H 0 2

8 76 X 1016 molecules H 0 2

imol

e.

molar mass of C H = 78.11 g 6 5.23 g x

6.022 x 10 molecules 22 = 4.03 x 10 molecules C H 6 78.llg

41.

a.

molar mass of C 0 = 46.07 g 6 H 2 1.27 1 g x

mol 46.07g

=

0.02759 mol C 0 6 H 2

0.02759molC 0 6 H 2 x b.

2molC =0.05518molC O 6 H 2 lmolC

molar mass of 2 C1 = 147.0 g 4 H 6 C 3.982 g x

1 mol 1470g

=

0.027088 mol C 2 C 4 H 6 1

0 027088 mol C 2 C 4 H 6 1x c

6 mol C 1 mol 2 C1 4 H 6 C

=

0 1625 mol C

molar mass of C 2 = 68 03 g 0 3 0.4438 g x

1 1 68.03 g

=

0.0065236 mol C 2 0 3

0.0065236 mol C 2 0 3 3 mol C 1 mol C 2 0 3

=

0.0 1957 mol C



ci.

C1 = 84.93 g CH molar mass of 2 2.910 g x

1 mol 84.93g

CI CH 0.034264 mol 2

C1 x CH 0.034264 mol 2

1 mci C C1, 2 1 mol CH

=

0.03426 mol C

42.

a.

SO = 142.1 g 2 Na molar mass of 4 2.01 g 4 SO x 2 Na

b.

SO = 126.1 g 2 Na molar mass of 3 SO x 2 gNa 2.01 3

c.

imolS =0.Ol4lmolS 4 S 2 lrnolNa O

4 S 2 lmolNa O 142.lg

3 S 2 O 1 molNa 126.1 g’

•

1 molS =0.Ol59molS SO 2 Na 1 mci 3

S = 78.05 g 2 molar mass of Na 2.OlgNa S 2 x

lmolNa S 2 imolS x —00258mo1S 1 molNa 78.05 g S 2 —

d.

3 = 158.1 g O S 2 molar mass of Na 2.01 gNa 3x O S 2

molNa 3 O S 12 I mci S < = 0.0127 mól S 158.1 g 1 molNa S 2

43. molar 44.

less than

45.

a.

mass of H present = 1.008 g = 1.008 g mass of Cl present = 35.45 g = 35.45 g mass of 0 present = 3(16.00 g) = 48.00 g 3 molar mass of HC1O

=

84.46 g

%H=

1.OO8gH xlOO=1.193%H 84.46 g

%Cl=

35.45 gCl xlOO=41.97%C1 84.46 g

%0=

48.OOgO xlOO=56.83%0 84.46 g

Copyright @ Houghton Mifflin Company. All rights reserved.

137

138


b.

mass ofU present = 238.0 g

=

238.0 g

mass ofF present = 4(19.00 g) = 76.00 g

molar mass of UF = 314.0 g 4

c.

%U=

238.OgU xlOO=75.80%U 314.Og

%F=

76.OOgF x100=24.20%F 314.0 g

mass of Ca present = 40.08 g = 40.08 g mass of H present = 2(1.008 g)

=

2.016 g

molar mass of CaH 2 = 42.10 g 40.O8gC xlOO=95.21%Ca 42.lOg

%Ca=

2.Ol6gH xlOO=4.789%H 42. 10 g

%H= d.

mass ofAg present =2(107.9 g)

=

215.8 g

mass of S present = 32.07 g = 32.07 g molar mass of Ag 5 2 %Ag=

=

247.9 g

215.8 gAg xlOO=87.06%Ag 247.9g

32.O7gS xlOO=12.94%S 247.9 g e.

mass of Na present = 22.99 g = 22.99 g mass of H present = 1.008 g = 1.008 g mass of S present = 32.07 g = 32.07 g mass of 0 present = 3(16.00 g) molar mass of NaHSO 3 % Na

=

=

=

48.00 g

104.07 g

22.99 g Na x 100 104.07 g

=

22.09% Na

%H

1.008 g H x 100 104.07 g

%S=

32.O7gS xIOO=30,82%S 104.07 g

%O=

48.OOgO xlOO=46.12%0 104.07 g

=

0.9686% H


Chapter 8: ChemIcal CompositIon f.

mass of Mn present =54,94

=

54.94 g

mass of 0 present = 2(16.00 g) = 32.00 g 2 = 6.94 g molar mass of Mn0 x 100 =63.19% Mn

86.94g

32.OOgO xlOO=36.81%O 86.94 g

%S= 46, a.

mass of Cu present = 2(63.55 g) = 127.1 g mass of 0 present = 16.00 g molar mass ofCu 0 = 143.1 g 2 %Cu=

127.1 CU % 1008882

143.1 g

% 1 0 6.OOgO xlOO=11.18%O 143.lg b.

mass of Cu present = 6355 g mass of 0 present = 16.00 g molar mass of CuO = 79.55 g % Cu

=

%0= c.

63.55 g Cu x 100 79.55g

=

79.89% Cu

16.OOgO x 100=20.11%0 79.55g

mass of Fe present = 55.85 g mass of 0 present = 16.00 g

V

molar mass of FeO = 71.85 g % Fe

=

=

55.85 g Fe 71.85 g 16.00 gO 71.85 g

X

100=77.73% Fe V

100

22.27% 0

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139

140

Chapter B: Chemical Composition

d.

mass ofFepresent=2(5.85 g)= 111.7g mass of 0 present = 3(16.00 g)

=

48.00 g

molar mass of Fe 3 = 159.7 g 0 2 %Fe= =

e.

111.7 gFe x 100=69.94% Fe 159.7g 48.00 gO 159.7 g

<

100

30.06% 0

mass of N present = 14.01 g mass of 0 present = 16.00 g molar mass ofNO = 30.01 g %N=

14.01 gN x 100 =46.68% N 30.01 g

%O=

16.00 gO 30.Olg

V

f.

V

100=53.32% 0

mass ofNpresent= 14.01g. mass of 0 present = 2(16.00 g) = 32.00 g

V

V

V

molar mass of NO 2 = 46.01 g V

%N=

14.01 gN x 100 =30.45% N 46.01 g

%O=

32.00 gO x 100=69.55% 0 46.01 g

V

V

47. a.

molar mass of CH 4 = 16.04 g. %C= xlOO=74.88%C 120 16.04 g

b.

molar mass NaNO 3 = 85.00 g 22.99gNa xl00=27.05%Na 85.00 g

%Na= c.

molar mass of CO %C=

d.

V

V

46.01 g

14.01 gN x 100 =30.45% N 46.01 g

V

ii

28.01 g

l2.OlgC xIOO=42.88%C 28.01 g

molar mass of NO 2 %N=

H.

V


• Chapter 8: ChemIcal Composition e.

molar mass of Q= 18 130.2 g H 8 C %C=

f.

molar mass of 4 (P0 =310.1 g 3 Ca 2 ) %Ca=

g.

120.2 gCa xlOO=38.76%Ca 310.1 g •

molar mass of 0 C 1 H 2 = 170.2 g 0 %C=

h.

96.O8gC xlOO=73.79%C 130.2 g

144.lgC xlOO=84.67%C 170.2 g

molar mass of A1(C ) = 204.1 g 0 3 H 2 %A1=

26.98gAI xlOO=13.22%Al 204J g

48. a.

molar mass of CuBr 2 = 223.4 %Cu

b.

molar mass of CuBr %Cu=

c

d.

55.85 gFe x10034.43%Fe 162.2g

58.93 gCo xlOO= 18.85% Co 312.7 g

molar mass of Co1 3 = 439.6 g %Co=

g.

55.85 gFe xlOO=44.06%Fe 126.75

molar mass of Co1 2 = 312.7 g %Co=

f.

63.55 gCu xlOO=44.29%Cu 143.5 g

molar mass of FeC1 3 = 162.2 g %Fe=

e.

143.5 g

molar mass of FeC1 2 = .126.75 g %Fe=

F

63.55 gCu xlOO=28.45%Cu 223.4g

58.93 g Co x10013.41%Co 439.6g

molar mass of SnO % Sn=

=

134.7 g

118.7 g Sn x 100=88.12% Sn 134.7g

Copyright © Houghton Mifflin Company, All rights reserved.

.4,,,

/ c. “

141

_____

• V4

142

44


h.

molar mass of Sn0 2

150.7 g V

150.7g 49. a.

molar mass of C 1 004 H 6 =

b.

100

molar mass ofNH 3 N 4 O %N=

c.

72.06 g C 146.1 g

=

146.1 g 49.32% C 80.05 g

28.O2gN xlOO=35.00%N 80.05 g

molar mass of CH 2 0 4 N 10

=

V

194.2 g

96.O9gC x 100=49.47%C l94.2g d.

molar mass of C10 2 %Cl=

V V

67.45 g

=

V

35.45 gCl x 100=52.56% Cl 67.45 g

V

V

V

V

e.

11 mo1armasso 0H=100.2g fCH %C=

V

72.O6gC xlOO=71.92%C 100.2g

V

V

V

f.

molar mass ofC 0 1 H 6 2 C

=

72.06 g C 180.2g

=

180.2 g

V

V

100 = 39.99% C

V

V

g

molar mass of 42 ()H 2 C %C=

h.

=

282 5 g

240.2gC xlOO=85.03%C 282.5g

molar mass of C OH 5 H 2

=

V

V

46.07 g

V

VV

V

V

46.07 g V

50. a

molar mass of FeC1 %Fe=

=

162 2 g

5585gFe x1003443%Fe 1622g

VV

Copyright © Houghton Mifflin Company All rights reserved.

Chapter 8: Chemlcal.Composltlan.

143

5

2 = 54.00 g molar mass of OF

b.

o

=

16.00 g 0 x 100 = 29.63% 0 54.00 .•

H = 78.11 g 6 molar mass of C

c.

=

72.06 g C 78.llg

100 = 92.25% C

CIO = 117.5 g NH molar mass of 4

d.

55

14.01 gN xlOO=11.92%N %N= 117.Sg

5

S

S

0=231.8g 2 molarmassofAg

e.

.

215.8gAg xlOO=93.10%Ag 231.8g

%Ag=

S

S S S

S

2 = 129.83 g molar mass of CoC1

f.

S

58.93 g Co 129.83g

% Co

X

100 = 45.39% Co

S

‘5

S

molar mass ofN 4 = 92.02 g 0 2

g.

%N=

S

S

28.O2gN x100=30.45%N 92.02g

S

S

S S

S S

h.

molar mass of MnC1 = 125.84 g 2 S

S

5

Mn 9 . 4 %s %fl 1004366 4 125.8g S

51. a.

molar mass of NHJ = 144.94 g 4.25gx

14 4 molNH I 1 molNFI =0.0293molNH. 14’.k94g I 4 lmolNH

S

S

molar mass of NFL = 18.04 g

b,

0.0293 mo1NH x 4

18.04 g NH 4 4 lmolNH

6.31 mol (NH S 2 ) 4

2 mol NH 4 = 12.6 mol NH S 2 ) 4 lmol(NH

=

S

0.529 gNH 4

S

S

S

SS

molar mass of NH = 18.04 g S

12.6 mol

18.04 g 4 lmolNH

=

S

227 g NH ion S

SS

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:

:

________

144

•

Chapter B Chemical Composition

c.

molar mass of Ba 2 = 473.8 g P 3 9.71 g

mol Ba 2 P 3 473.8g

<

3 mol Ba 2 2 P 3 lmolBa

=

0.0615 mol Ba 2 .

molar mass of Ba 2 = 137.3 g 137.3 g Ba 2 2 1mo1Ba

0.0615 mol Ba 2 • d.

(P0 3 Ca 2 ) 7.63 mol 4

=

844 g Ba 2

3 mol Ca 2 1 mol Ca 3 (P0 2 ) 4

=

22.9 mol Ca 2

molar mass of Ca 2 = 40.08 g 22.9 mol Ca 2x

40.08 g Ca 2 2 1mo1Ca

=

918 g Ca 2

52. a.

molar mass ofCl = 53.49 g; molar mass of M ion = 18.04 g % NH

=

18 04 g NH x 100 = 33 73% NH CJ 4 53.49gNH

•

“?

.

b.

molar mass of CuSO 4 = 159.62 g; molar mass of Cu 2 = 63.55 g 63.55 gCu 2 %Cu = 2 ‘159.62 g CuSO 4

• c.

2 100=39.81%Cu

molar mass of AuC1 3 = 303.4 g; molar mass of Au 3 ion = 197.0 g 3 % Au

d.

•.

=

197.0 g Au 3 3 x 100 = 64.93% Au 3 303.4gAuCl

molar mass of AgNO 3 = 169.9 g; molar mass of Ag ion = 107.9 g %Ag=

l07.9gAg 4 xlOO=63.5l%Ag 169.9 gAgNO 3

53.

To determine the empirical formula of a new compound, the composition of the compound by mass must be known. To determine the molecular formula of the compound, the molar mass of the compound must also be known.

54.

The empirical formula indicates the smallest whole number ratio of the number and type of atoms present in a molecule. For example, NO 2 and N 4 both have two oxygen atoms for every 0 2 nitrogen atom and therefore have the same empirical formula a.

NaO

b.

2 0 3 H 4 C

c.

C 1 H 3 0 2 N 2 is afrea4y the empirical formula.

d.

C1 3 H 2 C

Copyright @ Houghton Mifflin Company. All rights reserved,


145

56.

57.

a.

yes (Each of these has the empirical formula CH.)

b.

no (The number of hydrogen atoms is wrong.)

c.

.) 2 yes (Bpth have the empirical formula NO

d.

no (The number of hydrogen and oxygen atoms is wrong.)

Assume we havç 100.0 g of the compound so that the percentages become masses. 46.46 g Li x

11101 .6.941g

53.54 g 0 x

mci 16.OOg

6.694 mol Li

=

=

3.346 mol 0

Dividing both of these numbers of moles by the smaller number of moles (3.346 mol 0) gives 6.694 mol Li • 3.346 mol

2.000 mol Li

=

3.346mo10 =1.000molO 3,346 The empirical formula is Li 0 2 58.

Assume we have 100.0 g of the compound so that the percentages become masses. 98.55 g Ba x

1.447gHx

moi 137.3g •

=

imol 1008g

0.7178 mol Ba

1.4355molH

Dividing both of these numbers of moles by the §maller number of moles (0.7178 mol Ba) gives 0.7 178 mol Ba 0.7178

1.000molBa

1.4355moiH =2.000molH 0.7178mo1 •.

The empirical formula is Bali . 2 59.

0.2322 g C x

mol 12.Olg

0.05848 g H x 0.3091 gOx

=

mol 1.008g I mol 16.OOg

0.01933 mol C •.

0.05802 mol H 0.01932mo10


146

Chapter 8: ChamlcaE Composition

Dividing each number of moles by the smallest number of moles (0.01932 mol C) gives • 0.01933molC = 1.001 moiC 0.01932 mol• 0.05802mo1B =3.OO3molH 0.01932rno1 0.01932mo10 =1.000molO 0.01932mo1

•

•••.

S

The empirical formula is CH O. 3 60. Assume we have 100.0 g of the compound, so that.the percentages become masses. 1 mol 40.08g

28.03 g Ca x 22.38g0x 49.59 g Cl x

=

0;6994 mol Ca

•lmol =1.3988mo10 16.OOg mol 35.45 g

=

1.3989 mol Cl

rI

Dividing each number of moles by thesmallest number of moles (0.6994 mcii Ca) gives 0.6994 mol Ca =1.000molCa 06994 mol 1.3988mo10 0.6994 mol

2.000molO

S

1.3939 mol Cl • = 2.000 mol Cl 0.6994 mol The empirical formula is 2 C1 which is more commonly written as Ca(OCI) CaO , . 2 61.

The mass of chlorine in the reaction is 6.280 1.271 g Al x

mol 26.98 g

5.OO9gClx

Imol =0.l4l3molCl 35.45g

=

—

1.271

=

5.009 g Cl.

0.04711 mol Al

Dividing each of these number of moles by the smaller (0.04711 mol Al) gives 0.04711 mol Al = 1.000 mol Al 0.04711 mol 0.1413 mol Cl = 3.000 mol Cl 0.04711 mol The empirical formula is Aid . 3

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Chapter 8 ChemIcal CompositIon

147

V.

62. Consider 1O00 g of the compound. 1 mol 14.01 g

29.16 g N x

=

2.081 mol N V

8.392 g H x

1.008g

=

8.325 mcI H

12.5OgCx

imol =1.O4lmolC 12.01 g

49.95 g Ox

1 mol = 3.122 mcI 0 16.00g

V

V

V

4’. V V ‘V

Dividing each number of moles by the smallest (1.041 mol C) gives 2.081 tnolN 1.O4liiiol

i.999mo1N

V V

V

V

V

V

H 3 . 8 Smo 799 2 IHV 71 1.O4lmol

V •‘•V

V

1.O4lmoiC =1.000rnolC 1.O4lmol

VV1

V

V

3.l2lmolO =2.998rno10 1.041 mol V

The empirical formula is 3 C0 [i.e., ] 8 H 2 N 3 C 2 ) 4 (NH . 0

V

V

•V V

V

V

V

V

63.

1 mol 3.269 g Zn x 6.5.38g

V

0.05000 mol Zn

=

:.

V

V

V

1mo1 = 0.0500 mol 0 16.OOg

0.800 .g 0 x

V

V

V

V

V

Because the two components are present in e ual amounts on a molar basis, the empirical formula must be simply ZnO. V

64.

Consider 100.0 g of the compound. 55.06 g Co x

mci 58.93 g

0.9343 mol Co

If the sulfide of cobalt is 55.06% Co, then it is 44.94% S by mass. 44.94gSx

Imol =1.4OlmolS 32.07 g

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________= __________=

148

ChapterS: ChemIcal Composition Dividing each number of moles by the smaller (0.9343 mol Co) gives 0.09343 mol Co =1.000molCo 0.9343 l.4OlmolS 0.9343 mol

l.500mol S

Multiplying by two to.convert to whole numbers of moles gives the empirical formula for the . 3 S 2 compound as Co 65. The amount of fluorine that reacted with the aluminum sample must be (3.89 g 1.25 g) = 2.64 g of fluorine —

Imol =0.04633molAl 26.98g

1.25gAlx 2.64gFx

imol =0.l389molF 19.OOg

Dividing each number of moles by the smaller number of moles gives 0.04633 mol Al =1.000molAl 0.04633 mol O.1389molF =2.999molF 0.04633 mol . 3 The empirical formula is just AIF 1 aol = 0.09266 mol Al 66. 2.50 g Al x 26.98g 5.28 g F

x

1 mol = 0.2779 aol F 19.OOg

Dividing each number of moles by the smaller number of moles gives 0.09266 mol Al =l.000,moLAl 0.09266 mol 0.2779molF 0.09266 mol

2.999 aol F

. Note the similarity between this problem and question 65; they 3 The empirical formula is just AIF differ in the way the data is given. In question 65, you were given the mass of the product and first had to calculate how much fluorine had reacted. 67.

Consider 100.0 g of the compound. 67.6lgUx

Imol =0,284lmollJ 238.Og

32.39gFx

Imol 19.OOg

=

l.7O5molF



•

..

149

Dividing each number of moles by the smaller number of moles (0.2841 mol U) gives 0.2841 mol U l.00Omo1U 0.2841 mol 1.7O5molF =6.000molF 0.2841 mol The empirical formula is UF . 6

68. Consider 100.0 g of the compound.. 32.l3gAlx FS.

67.87 gFx

Imol =1.l9lmolAl 26.98g imol =3.572mo1F 19.OOg

Dividing each number ofmoles by the smaller number (1.191 mol Al) gives 1.191 mol Al =1.000molAl 1.l9lmol 3.572molF l.l9lmol

2.999 mol F

The empiric4l formula is AIF . Compare this question to questions 65 and 66: the three questions 3 illustrate the different forms in which data for calculating empirical formulas may occur. 69. Consider 100.0 g of the compound. 33.8 g 8 Cux irnol =0.533lmolCu 63.55g 14.94gNx 51.l8gOx

°

14.Olg

=1.O66molN

Imol =3.199mo10 16.OOg

Dividing each number ofmoles by the smallest number of moles (0.5331 mol Cu) gives 0.5331 mol Cu =1.000molCu 0.5331 mol 1.066 mol N • 0.5331 mol

=

2.000 mol N

3.199 mol 0 =6.OOlmolO 0.5331 mol The empirical formula is CuN 6 [i.e., 3 O 2 Cu(N0 ] 2 ) ,

Copyright © Houghton MIfflin Company. All rights reserved.

________ __________ ____________

150


70.

Consider 100.0 g of the compound. 59.78 g Li x

1 mol 6.941 g

40.22 g N x

1 mol 14.01 g

V•

8.6 13 mol Li

=

2.871 mol N

=

Dividing each number of moles by the smaller number of moles (2.871 mol N) gives 8.613 mol Li 2.871 mol 2.871 moIN 2.871 mol

3.000 mol Li

=

1.000molN

N. 3 The empirical formula is Li 71.

:

Consider 100.0 g of the compound. mol 63.55 g

66.75 g Cu x

=

1.050 mol Cu :.

V

V

I mol =0.500molP 10.84gPx 30.97 g 22.41 gOx

1 mol 16.OOg

=

V V

V

V

1.401 molO

V

Dividing each number of moles by the smallest number of moles (0.3500 mol P) gives V

1.050 mol Cu 0.3500 mol

=

3.000 mol Cu

0.3500molP 0 3500

=

1 000 mol P

V

1401 molO =4.OO3moIO 03500 mol

V.

V

V

V

V

O 4 P 3 Cu The empirical formula is thus . 72.

V

Consider 100.0 g of the compound. 15.77 g Al x 28.11 g S x 56.12 gO x

1 mol 26.98 g

V

=

0.5845 mol Al

V

V V

V V V

V

1 mol 32.07g 1 mol 16.OOg

V

VV

0.8765 mol S

V

V

V

V

V V

V

=

3.508 mol 0 V V


V

V

Chapter 8 Chemical Composition

151’

Dividing each number of moles by the smallest number of moles (0.5845 mol Al) gives 0.5845molAl =1.000molAl 0.5845mo1 0.8765 mol S =l.500molS 0.5845 mol 3.508mo10 0.5845rno1

6.002 mol 0

Multiplying these relative numbers of moles by two to give whole numbers gives the empirical 12 [i.e., 4 0 3 S 2 (S0 2 A1 j 3 ) . formula as A1 The compourid must contain 1.00 mg of lithium and 2.73 mg of fluorine.

73.

1.00mg Lix 2.73mgFx V

V

immol : = 0.144 mmol Li 6941mg .

V

immol =0.l44molF 19.00mg

The empirical formula of the compound is LiP. Compound 1: Assume 100.0 g of the compound.

74.

22.55gPx 7745 g Cl x

lVmol =0.728lmolP 30.97 g 1 mol 35.45g

=

.

V

V V

VVVVVV

V• V

2 185 mol Cl

Dividing each number of moles by the smaller (0.7281 mol P) indicates that the formula of . 3 Compound I is PCi

•

Compound 2: Assume 100.0 g of the compound. 14.87 g P x 85.l3gClx

1 mol 30.97g

=

V

V

0.4801 mol P V

imol =2.4OlmolCl 35.45g

V

Dividing each number of moles by the smaller (0.4801 mol P) indicates that the formula of Compound 2 is PCI . 5 The empirEcalformula of a compound represents only the smallest whole number relationship between the number and type of atoms in a compound, whereas the molecularformula represents the actual number of atoms of each type in a true molecule of the substance. Many compounds (for example, H 0) have the same empirical and molecular formulas. 2

75.

76. V

If only the empirical formula is known, the molar mass of the substance must be determined before the molecular formula can be calculated.

Copyright® Houghton Muffin Company All rights reserved. V

V

V

152

77.


Assume that we have 100.00 g of the compound; then 78.14 g will be boron, and 21.86 g will be hydrogen. 78.14 g B x

1 mol 10.81 g

21.86gHx

imol =21.69molH 1.008 g

=

7.228 mol B

Dividing each number of moles by the smaller number (7.228 mol B) gives the empirical formula 3 would be [10.81 g + 3(1.008 g)] = 13.83 g. This is as BH . The empirical molar mass of BH 3 approximately half of the indicated actual molar mass, and therefore the molecular formula must be B . 6 H 2 78.

empirical formula mass of CH = 13 g =

molar mass empirical formula mass

=

2!. 13 g

=

6

6 or C . H 6 The molecular formula is (CH) 79.

2 = 14 empirical formula mass of CH =


=

6

14 g

molecular formula is (CH 6=. ) 2 12 H 6 C 80.

empirical formular mass of C 0 5 H 2 =


=

46 g

=

46 g

=

2

molecular formula is 2 0) = 0 5 H (C 10 H 4 C 2 81.

Consider 100.0 g of the compound. 42.87gCx

imol =3.57OmolC 12.01 g

3.598 g H x

mol 1.008 g

=

3.569 mol H

28.55 g Ox

1 mol 16.OOg

=

1.784 mol 0

25.00 gNx

I mol 14.01 g

1.784molN



153

Dividing each number of moles by the smallest number of moles (1.784 mol 0 orN) gives. 3.57OmolC =2.OOlmolC 1.784mo1 3.569molH =.2.OOlmolH 1 784 mol l784mo10 100010 1.784mo1 l.784molN =l.000molN 1.784mo1 ON, and the empirical formula mass of C H 2 The empirical formula of the compound is C ON H 2 is 56.

molarmass empincal formula mass

168g =3 56 g

-

=

ON) H 2 (C The molecular formula is 3 82.

For NO , molar mass = 14.01 2 %N=

0

I4.01 N 100 46.01 g

+

=

. N 3 0 H 6 C

2(16.00) = 46,01 g.

3045°/N

2(16.00 g 0) x 100=69.55% 0 46.01 g

, molar mass =-2(14.0l g) + 4(16.00 g) 4 0 2 For N %N=

1003045 ( 2 . 4 % OlgN l 92.02 g

1006955 ( 4 . 6 0 OOO) l %O= % 92.02g 83. [1]

c

[6]

d

[2]

e

[7]

a

[3]

j

[8]

g

[4]

h

[9]

i

[5]

b

[10]

f

Copyright © Houghton Muffin Company All rights reserved.

=

92.02 g.

154


84. 5.OOgAl 0.140 g Fe gCu 2 2.7xlO 0.00250 g Mg 0.O62gNa gU 8 3.95 x 10

0.185mo1 0.00250 mol 4.3mol 1.03 x 10 mol 2.7x lO mol 3 1.66 x 10_20 mol

1.12x10 atoms 1.51 x 1021 atoms atoms 24 2.6xl0 6.19 x i0’ atoms 1.6x 1021 atoms 1.00 x i0 4 atoms

85. 4.24 g 4.04 g 1.98 g 45.9 g 126 g 0.297 g 86.

0.0543 mol 0.224 mol 0.0450 mol 1.26 mol 6.99 mol 0.00927 mol

3.27 1.35 2.71 7.59 4.21 5.58

x x x x x x

=

mass of I mol Y = 57.7 g

57.7 g

mass of 3 mol Z

=

3(63.9 g)

YZ 2 X molar mass of 3 %X

=

1022 molec.

1023 molec. 1024 molec. 1021 molec.

3.92 4.05 8.13 1.52 1.26 3.35

x x x x x x

1023 atoms 1023 atoms 1022 atoms 1024 atoms 1025 atoms 10 atoms

82.4 g

mass of2 mol X =2(41.2 g) =

1022 molec. 1023 molec.

191.7 g

331.8 g

82.4g xlOO=24.8%X 331.8 g 57.7g xlOO=17.4%Y 331.8 g

. % l 9 l 1 00578 %Z= 7 331.8 g , the percentage composition would be the same, 6 Z 2 Y 4 If the molecular formula were actually X and the relative mass of each element present would not change. The molecular formula is always a whole-number multiple of the empirical formula. 87.

magnesiumlnitrogen compound: mass of nitrogen 0.9240 g Mg x

0.3551 gN x

1 mol 24.31 g

1 mol 14.01 g

=

=

1.2791 g

—

0.9240

=

0.3551 g N

0.03801 mol Mg

0.02535 mol N

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Chemical Composition

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